To solve this exercise it is necessary to apply the equations related to the magnetic moment, that is, the amount of force that an image can exert on the electric currents and the torque that a magnetic field exerts on them.
The diple moment associated with an iron bar is given by,

Where,
Dipole momento associated with an Atom
N = Number of atoms
y previously given in the problem and its value is 2.8*10^{-23}J/T


The number of the atoms N, can be calculated as,

Where
Density
Molar Mass
A = Area
L = Length
Avogadro number


Then applying the equation about the dipole moment associated with an iron bar we have,



PART B) With the dipole moment we can now calculate the Torque in the system, which is



<em>Note: The angle generated is perpendicular, so it takes 90 ° for the calculation made.</em>
Answer: a = 1.32 * 10^18m/s² due north
Explanation: The magnitude of the force required to move the electron is given as
F = ma
The force exerted on the charge by the electric field of intensity (E) is given by
F = Eq
Thus
Eq = ma
a = E * q/ m
Where a = acceleration of charge
E = strength of electric field = 7400N/c
q = magnitude of electronic charge = 1.609 * 10^-6c
m = mass of an electronic charge = 9.109 * 10^-31kg
a = 7400 * 1.609 * 10^-16/ 9.109 * 10^-31
a = 11906.6 * 10^-16 / 9.019 * 10^-31
a = 1.19 * 10^-12 / 9.019 * 10^-31
a = 0.132 * 10^19
a = 1.32 * 10^18m/s²
As stated in the question, the direction of the electric field is due north hence, the direction of it force will also be north thus making the electron experience a force due north ( according to Newton second law of motion)
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